STAT 400

Mon. March 2nd, 2020

---

(Continuous) Cumulative Distribution Function

The cumulative distribution function for a continuous random variable $X$ with pdf $f(x)$ is $F(x)=P(X\le x)=\int_{-\infin}^x f(y)dy.$

Ex. $X$ follows $\text{unif}(A,B)$. Find the cdf of $X$.$\begin{aligned} f(x)&=\begin{cases} \frac{1}{B-A} & x\in[A,B]\\ 0 & \text{otherwise} \end{cases}\\\\ F(x)&=\begin{cases} 0 & x\le A\\ \int_{-\infin}^x f(y)dy=\int_A^x f(y)dy=\frac{X-A}{B-A} & A\le X\le B\\ \int_{-\infin}^x f(y)dy=\int_A^B f(y)dy = 1 & x\ge B \end{cases} \end{aligned}$

---

Properties

1. $F(x)$ is the total area under $f(y)$ to the left of $x$.
2. Since $f(y)\ge 0$, we know $F(x)$ is non-decreasing.
3. $0\le F(x)\le 1$.

Proposition. Given the pdf $f(x)$ and the cdf $F(x)$ of a continuous RV $X$, then: $\begin{aligned} &P(X>a)=1-P(X\le a)=1-F(a)\\ &P(a\le X\le b)=P(X\le b)-P(X\le a)=F(b)-F(a)\\ \end{aligned}$ Remember that since we’re dealing with continuous functions now, the edge cases of $X=a$ and $X=b$ don’t affect the cdf like they do for discrete variables.

Proposition. From the fundamental theorem of calculus, we have $f(x)=\frac{d}{dx}F(x).$ Thus you can calculate a pdf given a cdf and vice-versa.

---

Percentiles of a Continuous Distribution

In some cases we want to solve the inverse problem of finding the cdf $F(x)$ – that is, given $p$, we want to find $a$ such that $P(X\le a)=F(a)=p$.

Definition. Given $0\le p\le 1$, the $(100p)$th percentile of $X$, denoted as $\eta(p)$, satisfies $p=F(\eta(p))=\int_{-\infin}^{\eta(p)}f(x)dx.$

---

Ex. The pdf of $X$ is $f(x)=\begin{cases} \frac{1}{8}+\frac{3}{8}x & 0\le x\le 2\\ 0 &\text{otherwise} \end{cases}$ and its cdf is $F(X)=\begin{cases} 0&x\le 0\\ \frac{x}{8}+\frac{3}{16}x^2&0\le x\le 2\\ 1&x>2 \end{cases}$

Given $p=\frac{1}{16}$ find $\eta(p)$. $\begin{aligned} F(\eta(p))&=\frac{1}{16}\\ \frac{\eta(p)}{8}+\frac{3}{16}\eta(p)^2&=\frac{1}{16}\\ \eta(p)&=-1\text{ or }\frac{1}{3}\\ \eta(p)&=\frac{1}{3}. \end{aligned}$

---

Median

Definition. For $p=0.5$, we call $\eta(0.5)$ the median, denoted by $\tilde{\mu}$. By definition, $F(\tilde{\mu})=0.5$.

Remark. $\mu=\tilde{\mu}$ when the pdf being studied is symmetric; i.e. $f(\tilde{\mu}+x)=f(\tilde{\mu}-x)$ for all $x$.
For instance, if $X$ follows $\text{unif}(A,B)$, $\tilde{\mu}=\mu=\frac{A+B}{2}.$

---

Expected Value

Definition. The expected value of a continuous random variable $X$ with pdf $f(x)$ is $\mu=\mu_X=E(X)=\int_{-\infin}^{\infin}xf(x)dx.$

Ex. $X$ follows $\text{unif}(A,B)$. $\begin{aligned} E(X)&=\int_A^Bx\cdot \frac{1}{B-A}dx\\ &=\frac{1}{B-A}\cdot \frac{B^2-A^2}{2}\\ &=\frac{(B+A)(B-A)}{2(B-A)}\\ &=\frac{B+A}{2}. \end{aligned}$ This result is in line with our remark about the median of random variables with symmetric pdfs.

---

Proposition. If $X$ is a continuous RV with pdf $f(x)$, then the expected value of $h(X)$ is $\mu_{h(X)}=E(h(X))=\int_{-\infin}^{\infin}h(x)f(x)dx.$

Ex. $X$ follows $\text{unif}(A,B)$. Find $E(X^2)$.
$\begin{aligned} E(X^2)&=\int_{-\infin}^{\infin}x^2f(x)dx\\ &=\int_A^Bx^2\cdot\frac{1}{B-A}dx\\ &=\frac{1}{B-A}\cdot \frac{B^3-A^3}{3}\\ B^3-A^3&=(B-A)(B^2+BA+A^2)\\ \rightarrow E(X^2)&=\frac{1}{3}(B^2+BA+A^2). \end{aligned}$

---

Variance and Standard Deviation

Definition. The variance of a continuous random variable $X$ with pdf $f(x)$ and expected value $\mu$ is $\sigma_X^2=V(X)=\int_{-\infin}^{\infin}(x-\mu)^2f(x)dx=E((X-\mu)^2).$
The standard deviation (SD) of $X$ is $\sigma_X=\sqrt{V(X)}$.

Proposition. $V(X)=E(X^2)-[E(X)]^2$, exactly the same property as with discrete RVs.

Ex. $X$ follows $\text{unif}(A,B)$. $V(X)=\frac{1}{3}(B^2+BA+A^2)-(\frac{B+A}{2})^2=\frac{1}{12}(A-B)^2.$